A Mechanical Analog
Sum of the Forces
RF = mg - bv = ma
Solve for the acceleration
b
dv
g- mv = a =
dt
Separate Variables
dv
= dt
b
g- mv
Integrate
#
v
0
-m
b
[ln (g - m v) - ln (g)] = t
b
dv
=
b
g- mv
t
dt
0
b
(g - m v) /g = e-bt/m
take exponential of both sides
v (t) =
#
mg
(1 - e-bt/m)
b
Physics 196 Chapter 31 Problem 66
The bar of mass m in is pulled horizontally across parallel rails by a massless string that passes over an ideal
pulley and is attached to a suspended mass M. The uniform magnetic field has a magnitude B, and the
distance between the rails is . The rails are connected at one end by a load resistor R. Derive an expression
that gives the horizontal speed of the bar as a function of time.
Induced current and
Magnetic Field
FreeBody Diagrams
y
FB
a
FT
y
v
a
FN
FB
FT
x
M-block
Bext
Fg
Faradayʼs Law
f = -dU
dt
x
m-bar
Fg
FB = BIL
B
The magnetic flux is changing through the area bounded by the bar and the rails. According to Faradayʼs Law a
changing magnetic flux will induce an emf in the circuit which will then produce a current which will then produce an
induced magnetic field that will oppose the changing magnetic flux.
Looking at the diagram, the direction of the induced current is counterclockwise. This will produce a magnetic field
that will add to the external magnetic field through the circuit. Since the arrows are pointing in the same direction
there is a repulsion between the induced and the external magnetic fields to the right.
Since the magnetic flux is decreasing through the circuit the induced magnetic field will produce a force on the bar
that will oppose this change.
Sum of the Forces
RFy = FT - Fg =-Ma y
RFx = FB - FT =-ma x
ay = ax
Falling Mass
FT = mg - Ma
Sliding Bar
FT = FB + ma
Solve for the acceleration
We need to find the induced magnetic force
mg - Ma = FB + ma
UB = BA cos i
Iind
f = dU
=
R
d (BA) 1
d (Blw) 1
1
dw 1
Blv
=
=
= Bl
=
R
dt R
dt R
dt R
dt R
mg - FB = Ma + ma
B
2 2
FB = BIind L = B (
Blv
Blv
)l =
R
R
mg - FB
=a
M+m
mg - FB
dv
=
M+m
dt
Now, we are set to solve for the velocity of the
bar with respect to time
2 2
mg
Blv
dv
+
=
M+m
dt
R (M + m)
Letʼs reduce our stress and set
mg
a= M+m
b=
Separate Variables
2 2
Bl
R (M + m)
dv
mg
B 2l 2 v
+
M+m
R (M + m)
= dt
Integrate
#
0
v
dv
=
a - bv
#
t
dt
0
-1
[ln (a - bv) - ln a] = t
b
ln (a - bv) - ln a =-bt
(a - bv) /a = e-bt
v (t) =
a
(1 - e-bt)
b
Over time, this magnetic force will balance with the gravitational force on the bar. This magnetic force is dependent
upon the velocity of the bar. The bar will reach a terminal velocity once these two forces acting on the bar are equal.
A Mechanical Analog
Sum of the Forces
RF = mg - bv = ma
Solve for the acceleration
b
dv
g- mv = a =
dt
Separate Variables
dv
= dt
b
g- mv
#
0
Integrate
v
dv
=
b
g- mv
#
t
dt
0
2 pts
b
(g - m v) /g = e-bt/m
2 pts
Solve for Velocity
Physics 196 Chapter 31 Problem 66
The bar of mass m in is pulled horizontally across parallel rails by a massless string that passes over an ideal
pulley and is attached to a suspended mass M. The uniform magnetic field has a magnitude B, and the
distance between the rails is . The rails are connected at one end by a load resistor R.
Derive an expression that gives the horizontal speed of the bar as a function of time.
Draw Induced current, Magnetic
Field and Mag-Force.
FreeBody Diagrams
2 pts
3 pts
y
a
Bext
FT
x
M-block
Fg
Faradayʼs Law
1 pts
The magnetic flux is changing through the area bounded by the bar and the rails. According to Faradayʼs
Law a changing magnetic flux will induce an emf in the circuit which will then produce a current which will
then produce an induced magnetic field that will oppose the changing magnetic flux.
Looking at the diagram, the direction of the induced current is counterclockwise. This will produce a
magnetic field that will add to the external magnetic field through the circuit. Since the arrows are pointing in
the same direction there is a repulsion between the induced and the external magnetic fields to the right.
Since the magnetic flux is decreasing through the circuit the induced magnetic field will produce a force on
the bar that will oppose this change.
Sum of the Forces
RFy = FT - Fg =-Ma y
1 pts
ay = ax
Falling Mass
Sliding Bar
FT = mg - Ma
1 pts
We need to find the induced magnetic force
Set eqʼs equal and solve for the acceleration
UB = BA cos i
Iind
2 pts
2 pts
f=
=
R
2 pts
mg - FB
=a
M+m
mg - FB
dv
=
M+m
dt
Plug in FB and simplify.
FB = BIind L =
2 pts
Letʼs reduce our stress and set
mg
a= M+m
b=
2 2
Bl
R (M + m)
Separate Variables
2 pts
Draw the graph of the velocity wrt time. What is the
terminal velocity? Show on the graph the velocity at e-2?
4 pts
Integrate
#
0
v
dv
=
a - bv
#
t
dt
0
2 pts
(a - bv) /a = e-bt
Solve for Velocity
2 pts